11) A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answers (1)

Let l and bare the length and breadth of rectangle respectively and r will be  the radius of circle (r = \frac{l}{2})
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle

                                          = l+2b + \pi \frac{l}{2}

                                         
l+2b + \pi \frac{l}{2} = 10\\ l = \frac{2(10-2b)}{2+\pi}
Area of window id given by (A) = lb + \frac{\pi}{2}\left ( \frac{l}{2} \right )^2  
                                                 = \frac{2(10-2b)}{2+\pi}b + \frac{\pi}{2}\left ( \frac{10-2b}{2+\pi} \right )^2\\
A^{'}(b) = \frac{20-8b}{2+\pi}+\frac{\pi}{2}.2(\frac{10-2b}{2+\pi}).\frac{(-2)}{2+\pi}
             = \frac{20-8b}{2+\pi}-2\pi(\frac{10-2b}{(2+\pi)^2})\\ A^{'}(b) = 0\\ \frac{20-8b}{2+\pi}=2\pi(\frac{10-2b}{(2+\pi)^2})\\ 40 + 20\pi -16b -8\pi b = 20\pi - 4\pi b\\ 40 = 4b(\pi+4)\\b = \frac{10}{\pi+4}
Now,
A^{''}(b) = \frac{-8}{2+\pi}+\frac{4\pi}{(2+\pi)^2} = \frac{-16-8\pi+4\pi}{(2+\pi)^2} = \frac{-16-4\pi}{(2+\pi)^2} \\ A^{''}(\frac{10}{\pi+4}) < 0
Hence, b = 5/2 is the point of maxima
l = \frac{2(10-2b)}{2+\pi} = \frac{2(10-2.\frac{10}{4+\pi})}{2+\pi} = \frac{20}{4+\pi}
r= \frac{l}{2}= \frac{20}{2(4+\pi)}=\frac{10}{4+\pi}
Hence, these are the  dimensions of the window to admit maximum light through the whole opening

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